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Math Help - cyclic quadrilateral question

  1. #1
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    cyclic quadrilateral question

    help me solve the attached question.
    Attached Thumbnails Attached Thumbnails cyclic quadrilateral question-cyclic-quad.jpg  
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  2. #2
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    Quote Originally Posted by mathkidx View Post
    help me solve the attached question.
    Do you know these?
    1) The measure of an inscribed angle in a circle is equal to half the measure of the intercepted arc.
    2) The measure of the angle made by a tangent line and a secant is half the meaure of the intercepted arc.
    3) The base angles of an isosceles triangle are congruent or equal in measures.
    4) The 3 interior angles of a triangle add up to 180 degrees.
    5) The opposite interior angles of a cyclic quadrilateral are supplementary, or add up to 180 degrees.

    If you do, then let us go to the diagram.

    If KN = KL, then triangle NKL is isosceles. Then the base angles KNL and KLN are equal. Let us call each as angle theta or t.

    Then, minor arc NK = 2(t).
    Also, minor arc KL = 2(t).

    ---------------------------
    To solve for angle KLN:

    angle PNL = (1/2)[minor arc NK + minor arc KL) = (1/2)(2t +2t) = 2t
    angle PNL = 24deg +t .......as shown on the diagram
    so,
    2t = 24deg +t
    t = 24 degrees -------**

    So, since angle KLN is t, then angle KLN = 24 degrees. -----answer.

    --------------------
    To solve for angle NKL:

    angle NKL = 180deg -t -t
    angle NKL = 180 -24 -24 = 132 degrees -------answer.

    -------------------------
    To solve for angle LMN:

    Angle LMN is the supplement of angle NKL. so,
    angle LMN = 180 -132 = 48 degrees. ------------answer.

    We can check that in the diagram,
    angle LMN = (1/2)[minor arc NK + minor arc KL)
    angle LMN = (1/2)(2t +2t) = 2t == 2(24) = 48 degrees.
    Last edited by ticbol; September 2nd 2008 at 05:22 PM.
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